"""
难度：简单
定义栈的数据结构，请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中，
调用 min、push 及 pop 的时间复杂度都是 O(1)。
示例:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min();   --> 返回 -3.
minStack.pop();
minStack.top();      --> 返回 0.
minStack.min();   --> 返回 -2.
提示：
各函数的调用总次数不超过 20000 次"""
class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
    class Node:
        def __init__(self,x):
            self.val = x
            self.min = x
    def push(self, x: int) -> None:
        node = self.Node(x)
        if len(self.stack) == 0:
            self.stack.append(node)
        else:
            topNode = self.stack[-1]
            if node.min > topNode.min:
                node.min = topNode.min
            self.stack.append(node)
    def pop(self) -> None:
        self.stack.pop()
    def top(self) -> int:
        return self.stack[-1].val
    def min(self) -> int:
        return self.stack[-1].min



# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()